Question

If $y^2 = ax^2 + bx + c$, then $\frac{d}{dx}\left(y^{3}\,y_{2}\right)=$

• A. $1$
• B. $-1$
• C. $\frac{4ac-b^{2}}{a^{2}}$
• D. $0$

Answer 1

Answer: (D)

Given $y^2 = ax^2 + bx + c$
Differentiating both sides, we get $2yy_1 = 2ax + b \quad \ldots(i)$
Again differentiating, we get $2yy_2 + y_1(2y_1) = 2a$
$\Rightarrow yy_{2}=a-y^{2}_{1}$
$\Rightarrow yy_{2}=a-\left(\frac{2ax+b}{2y}\right)^{2}$ (using $(i)$)
$=\frac{4y^{2}\,a-\left(4a^{2}\,x^{2}+b^{2}+4abx\right)}{4y^{2}}$
$\Rightarrow y^{3}\,y_{2}=\frac{4a\left(ax^{2}+bx+c\right)-\left(4a^{2}\,x^{2}+b^{2}+4abx\right)}{4}$
$=\frac{4ac-b^{2}}{4}$
$\Rightarrow \, \frac{d}{dx}\left(y^{3}\,y_{2}\right)=0$